The upper frac{3}{4}th portion of a vertical | vedclass

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Question

(B) Let the total height of the pole be $h$. The pole is divided into two parts: the upper part of length $\frac{3h}{4}$ and the lower part of length

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let the total height of the pole be $h$. The pole is divided into two parts: the upper part of length $\frac{3h}{4}$ and the lower part of length $\frac{h}{4}$.Let the point on the ground be $C$ and the foot of the pole be $B$. Given $BC = 40 \ m$.Let $	heta$ be the angle of elevation of the top of the pole $(A)$ and $\alpha$ be the angle of elevation of the point $D$ (where the pole is divided).Then $	an 	heta = \frac{h}{40}$ and $	an \alpha = \frac{h/4}{40} = \frac{h}{160}$.The angle subtended by the upper part $AD$ at $C$ is $\beta = 	heta - \alpha = 	an^{-1}\left(\frac{3}{5}ight)$.Thus,$	an \beta = \frac{	an 	heta - 	an \alpha}{1 + 	an 	heta 	an \alpha} = \frac{3}{5}$.Substituting the values: $\frac{\frac{h}{40} - \frac{h}{160}}{1 + \left(\frac{h}{40}ight)\left(\frac{h}{160}ight)} = \frac{3}{5}$.$\frac{\frac{3h}{160}}{1 + \frac{h^2}{6400}} = \frac{3}{5}$ $\Rightarrow \frac{3h}{160} 	imes \frac{6400}{6400 + h^2} = \frac{3}{5}$.$\frac{120h}{6400 + h^2} = \frac{3}{5}$ $\Rightarrow 600h = 3(6400 + h^2)$ $\Rightarrow 200h = 6400 + h^2$.$h^2 - 200h + 6400 = 0 \Rightarrow (h - 160)(h - 40) = 0$.So,$h = 40 \ m$ or $h = 160 \ m$.

The upper $\frac{3}{4}$th portion of a vertical pole subtends an angle $\tan^{-1}\left(\frac{3}{5}\right)$ at a point in the horizontal plane through its foot and at a distance $40 \ m$ from the foot. $A$ possible height of the vertical pole is $....... \ m$.

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